Proof. (a) Choose a total set \(\lbrace \varphi _n:n\in \mathbb {N}\rbrace \)  of unit vectors in \(\mathcal {H}\) . Then the probability measure \(\mu \)  on the Borel sets of \(\varepsilon \)  defined by \(\mu (\Delta ):=\sum _n2^{-n}\langle \varphi _n,T(\Delta )\varphi _n\rangle \)  has the same null sets as \(T\) , since generally \(\langle \varphi ,T(\Delta )\varphi \rangle =||\sqrt{T(\Delta )}\varphi ||^2=0\)  if and only if \(T(\Delta )\varphi =0\) .
So \(\mu \)  is quasi-invariant under translations \(b\in \varepsilon \) , since due to translational covariance of \(T\)  (see Eq. (REF )) one has \(\mu (b+\Delta )=\sum _n2^{-n}\langle U(b)^{-1}\varphi _n,T(\Delta )U(b)^{-1}\varphi _n\rangle =0\)  if and only if \(T(\Delta )=0\) . Hence \(\mu \)  is equivalent to the Lebesgue measure (see e.g. [1]}).
